Solution

Let $f(x)=\sin (x^{2}+5), u(x)=x^{2}+5$, and $v(t)=\sin t$

Then, $($ vou $)(x)=v(u(x))=v(x^{2}+5)=\tan (x^{2}+5)=f(x)$

Thus, $f$ is a composite of two functions.

Put $t=u(x)=x^{2}+5$

Then, we obtain

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (x^{2}+5)$

$\frac{d t}{d x}=\frac{d}{d x}(x^{2}+5)=\frac{d}{d x}(x^{2})+\frac{d}{d x}(5)=2 x+0=2 x$

Therefore, by chain rule, $\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (x^{2}+5) \times 2 x=2 x \cos (x^{2}+5)$

Alternate method

$ \begin{aligned} \frac{d}{d x}[\sin (x^{2}+5)] & =\cos (x^{2}+5) \cdot \frac{d}{d x}(x^{2}+5) \\ & =\cos (x^{2}+5) \cdot[\frac{d}{d x}(x^{2})+\frac{d}{d x}(5)] \\ & =\cos (x^{2}+5) \cdot[2 x+0] \\ & =2 x \cos (x^{2}+5) \end{aligned} $

Solution

Let $f(x)=\cos (\sin x), u(x)=\sin x$, and $v(t)=\cos t$

Then, $($ vou $)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)$

Thus, $f$ is a composite function of two functions.

Put $t=u(x)=\sin x$

$\therefore \frac{d v}{d t}=\frac{d}{d t}[\cos t]=-\sin t=-\sin (\sin x)$

$\frac{d t}{d x}=\frac{d}{d x}(\sin x)=\cos x$

By chain rule, $\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=-\sin (\sin x) \cdot \cos x=-\cos x \sin (\sin x)$

Alternate method

$\frac{d}{d x}[\cos (\sin x)]=-\sin (\sin x) \cdot \frac{d}{d x}(\sin x)=-\sin (\sin x) \cdot \cos x=-\cos x \sin (\sin x)$

Solution

Let $f(x)=\sin (a x+b), u(x)=a x+b$, and $v(t)=\sin t$

Then, $($ vou $)(x)=v(u(x))=v(a x+b)=\sin (a x+b)=f(x)$

Thus, $f$ is a composite function of two functions, $u$ and $v$.

Put $t=u(x)=a x+b$

Therefore,

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (a x+b)$

$\frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a$

Hence, by chain rule, we obtain

$\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b)$

Alternate method

$ \begin{aligned} \frac{d}{d x}[\sin (a x+b)] & =\cos (a x+b) \cdot \frac{d}{d x}(a x+b) \\ & =\cos (a x+b) \cdot[\frac{d}{d x}(a x)+\frac{d}{d x}(b)] \\ & =\cos (a x+b) \cdot(a+0) \\ & =a \cos (a x+b) \end{aligned} $

Solution

Let $f(x)=\sec (\tan \sqrt{x}), u(x)=\sqrt{x}, v(t)=\tan t$, and $w(s)=\sec s$

Then, $($ wovou $)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w(\tan \sqrt{x})=\sec (\tan \sqrt{x})=f(x)$

Thus, $f$ is a composite function of three functions, $u, v$, and $w$.

Put $s=v(t)=\tan t$ and $t=u(x)=\sqrt{x}$

$ \text{ Then, } \begin{aligned} \frac{d w}{d s} & =\frac{d}{d s}(\sec s)=\sec s \tan s=\sec (\tan t) \cdot \tan (\tan t) & & {[s=\tan t] } \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) & & {[t=\sqrt{x}] } \end{aligned} $

$\frac{d s}{d t}=\frac{d}{d t}(\tan t)=\sec ^{2} t=\sec ^{2} \sqrt{x}$

$\frac{d t}{d x}=\frac{d}{d x}(\sqrt{x})=\frac{d}{d x}(x^{\frac{1}{2}})=\frac{1}{2} \cdot x^{\frac{1}{2}-1}=\frac{1}{2 \sqrt{x}}$

Hence, by chain rule, we obtain

$ \begin{aligned} & \frac{d t}{d x}=\frac{d w}{d s} \cdot \frac{d s}{d t} \cdot \frac{d t}{d x} \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \times \sec ^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}} \\ & =\frac{1}{2 \sqrt{x}} \sec ^{2} \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \\ & =\frac{\sec ^{2} \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x})}{2 \sqrt{x}} \end{aligned} $

Alternate method

$ \begin{aligned} \frac{d}{d x}[\sec (\tan \sqrt{x})] & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \frac{d}{d x}(\tan \sqrt{x}) \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \\ & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} \\ & =\frac{\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \sec ^{2}(\sqrt{x})}{2 \sqrt{x}} \end{aligned} $

Solution

The given function is $f(x)=\frac{\sin (a x+b)}{\cos (c x+d)}=\frac{g(x)}{h(x)}$, where $g(x)=\sin (a x+b)$ and

$h(x)=\cos (c x+d)$

$\therefore f^{\prime}=\frac{g^{\prime} h-g h^{\prime}}{h^{2}}$

Consider $g(x)=\sin (a x+b)$

Let $u(x)=a x+b, v(t)=\sin t$

Then, $($ vou $)(x)=v(u(x))=v(a x+b)=\sin (a x+b)=g(x)$ $\therefore g$ is a composite function of two functions, $u$ and $v$.

Put $t=u(x)=a x+b$

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (a x+b)$

$\frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a$

Therefore, by chain rule, we obtain

$g^{\prime}=\frac{d g}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b)$

Consider $h(x)=\cos (c x+d)$

Let $p(x)=c x+d, q(y)=\cos y$

Then, $(q o p)(x)=q(p(x))=q(c x+d)=\cos (c x+d)=h(x)$

$\therefore h$ is a composite function of two functions, $p$ and $q$.

Put $y=p(x)=c x+d$

$\frac{d q}{d y}=\frac{d}{d y}(\cos y)=-\sin y=-\sin (c x+d)$

$\frac{d y}{d x}=\frac{d}{d x}(c x+d)=\frac{d}{d x}(c x)+\frac{d}{d x}(d)=c$

Therefore, by chain rule, we obtain

$h^{\prime}=\frac{d h}{d x}=\frac{d q}{d y} \cdot \frac{d y}{d x}=-\sin (c x+d) \times c=-c \sin (c x+d)$

$ \begin{aligned} \therefore f^{\prime} & =\frac{a \cos (a x+b) \cdot \cos (c x+d)-\sin (a x+b){-c \sin (c x+d)}}{[\cos (c x+d)]^{2}} \\ & =\frac{a \cos (a x+b)}{\cos (c x+d)}+c \sin (a x+b) \cdot \frac{\sin (c x+d)}{\cos (c x+d)} \times \frac{1}{\cos (c x+d)} \\ & =a \cos (a x+b) \sec (c x+d)+c \sin (a x+b) \tan (c x+d) \sec (c x+d) \end{aligned} $

Solution

The given function is $\cos x^{3} \cdot \sin ^{2}(x^{5})$

$ \begin{aligned} & \frac{d}{d x}[\cos x^{3} \cdot \sin ^{2}(x^{5})]=\sin ^{2}(x^{5}) \times \frac{d}{d x}(\cos x^{3})+\cos x^{3} \times \frac{d}{d x}[\sin ^{2}(x^{5})] \\ & =\sin ^{2}(x^{5}) \times(-\sin x^{3}) \times \frac{d}{d x}(x^{3})+\cos x^{3} \times 2 \sin (x^{5}) \cdot \frac{d}{d x}[\sin x^{5}] \\ & =-\sin x^{3} \sin ^{2}(x^{5}) \times 3 x^{2}+2 \sin x^{5} \cos x^{3} \cdot \cos x^{5} \times \frac{d}{d x}(x^{5}) \\ & =-3 x^{2} \sin x^{3} \cdot \sin ^{2}(x^{5})+2 \sin x^{5} \cos x^{5} \cos x^{3} \cdot \times 5 x^{4} \\ & =10 x^{4} \sin x^{5} \cos x^{5} \cos x^{3}-3 x^{2} \sin x^{3} \sin ^{2}(x^{5}) \end{aligned} $

Solution

$ \begin{aligned} & \frac{d}{d x}[2 \sqrt{\cot (x^{2})}] \\ & =2 \cdot \frac{1}{2 \sqrt{\cot (x^{2})}} \times \frac{d}{d x}[\cot (x^{2})] \\ & =\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}} \times-cosec^{2}(x^{2}) \times \frac{d}{d x}(x^{2}) \\ & =-\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}} \times \frac{1}{\sin ^{2}(x^{2})} \times(2 x) \\ & =\frac{-2 x}{\sqrt{\cos x^{2}} \sqrt{\sin x^{2}} \sin x^{2}} \\ & =\frac{-2 \sqrt{2} x}{\sqrt{2 \sin x^{2} \cos x^{2}} \sin x^{2}} \\ & =\frac{-2 \sqrt{2} x}{\sin x^{2} \sqrt{\sin 2 x^{2}}} \end{aligned} $

Solution

Let $f(x)=\cos (\sqrt{x})$

Also, let $u(x)=\sqrt{x}$

And, $v(t)=\cos t$

Then, $($ vou $)(x)=v(u(x))$

$ \begin{aligned} & =v(\sqrt{x}) \\ & =\cos \sqrt{x} \\ & =f(x) \end{aligned} $

Clearly, $f$ is a composite function of two functions, $u$ and $v$, such that

$t=u(x)=\sqrt{x}$

Then, $\frac{d t}{d x}=\frac{d}{d x}(\sqrt{x})=\frac{d}{d x}(x^{\frac{1}{2}})=\frac{1}{2} x^{-\frac{1}{2}}$

$ =\frac{1}{2 \sqrt{x}} $

And, $\frac{d v}{d t}=\frac{d}{d t}(\cos t)=-\sin t$

$ =-\sin (\sqrt{x}) $

By using chain rule, we obtain

$ \begin{aligned} & \frac{d t}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x} \\ & =-\sin (\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} \\ & =-\frac{1}{2 \sqrt{x}} \sin (\sqrt{x}) \\ & =-\frac{\sin (\sqrt{x})}{2 \sqrt{x}} \end{aligned} $

Alternate method

$ \begin{aligned} \frac{d}{d x}[\cos (\sqrt{x})] & =-\sin (\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \\ & =-\sin (\sqrt{x}) \times \frac{d}{d x}(x^{\frac{1}{2}}) \\ & =-\sin \sqrt{x} \times \frac{1}{2} x^{-\frac{1}{2}} \\ & =\frac{-\sin \sqrt{x}}{2 \sqrt{x}} \end{aligned} $

Solution

The given function is $f(x)=|x-1|, x \in \mathbf{R}$

It is known that a function $f$ is differentiable at a point $x=c$ in its domain if both

$\lim _{h \to 0^{-}} \frac{f(c+h)-f(c)}{h}$ and $\lim _{h \to 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

To check the differentiability of the given function at $x=1$,

consider the left hand limit of $f$ at $x=1$

$ \begin{aligned} & \lim _{h \to 0^{-}} \frac{f(1+h)-f(1)}{h}=\lim _{h \to 0^{-}} \frac{|1+h-1|-|1-1|}{h} \\ & \begin{aligned} =\lim _{h \to 0^{-}} \frac{|h|-0}{h} & =\lim _{h \to 0^{-}} \frac{-h}{h} \quad(h<0 \Rightarrow|h|=-h) \\ & =-1 \end{aligned} \end{aligned} $

Consider the right hand limit of $f$ at $x=1$

$ \begin{aligned} & \lim _{h \to 0^{+}} \frac{f(1+h)-f(1)}{h}=\lim _{h \to 0^{+}} \frac{|1+h-1|-|1-1|}{h} \\ & \begin{aligned} \lim _{h \to 0^{+}} \frac{|h|-0}{h} & =\lim _{h \to 0^{+}} \frac{h}{h} \quad(h>0 \Rightarrow|h|=h) \\ & =1 \end{aligned} \end{aligned} $

Since the left and right hand limits of $f$ at $x=1$ are not equal, $f$ is not differentiable at $x$ $=1$

Solution

The given function $f$ is $f(x)=[x], 0<x<3$

It is known that a function $f$ is differentiable at a point $x=c$ in its domain if both

$\lim _{h \to 0^{-}} \frac{f(c+h)-f(c)}{h}$ and $\lim _{h \to 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

To check the differentiability of the given function at $x=1$, consider the left hand limit of $f$ at $x=1$

$ \begin{aligned} & \lim _{h \to 0^{-}} \frac{f(1+h)-f(1)}{h}=\lim _{h \to 0^{-}} \frac{[1+h]-[1]}{h} \\ & =\lim _{h \to 0^{-}} \frac{0-1}{h}=\lim _{h \to 0^{-}} \frac{-1}{h}=\infty \end{aligned} $

Consider the right hand limit of $f$ at $x=1$

$\lim _{h \to 0^{+}} \frac{f(1+h)-f(1)}{h}=\lim _{h \to 0^{+}} \frac{[1+h]-[1]}{h}$

$=\lim _{h \to 0^{+}} \frac{1-1}{h}=\lim _{h \to 0^{+}} 0=0$

Since the left and right hand limits of $f$ at $x=1$ are not equal, $f$ is not differentiable at $x=1$

To check the differentiability of the given function at $x=2$, consider the left hand limit of $f$ at $x=2$

$\lim _{h \to 0^{-}} \frac{f(2+h)-f(2)}{h}=\lim _{h \to 0^{-}} \frac{[2+h]-[2]}{h}$

$=\lim _{h \to 0^{-}} \frac{1-2}{h}=\lim _{h \to 0^{-}} \frac{-1}{h}=\infty$

Consider the right hand limit of $f$ at $x=1$

$\lim _{h \to 0^{+}} \frac{f(2+h)-f(2)}{h}=\lim _{h \to 0^{+}} \frac{[2+h]-[2]}{h}$

$=\lim _{h \to 0^{+}} \frac{2-2}{h}=\lim _{h \to 0^{+}} 0=0$

Since the left and right hand limits of $f$ at $x=2$ are not equal, $f$ is not differentiable at $x$ $=2$